Why do decision-tree models not require feature scaling?

Feature scaling is needed because raw features often differ in units and magnitudes, making them incomparable. Scaling brings them onto a common scale, which is necessary for many machine learning algorithms to function properly.

Conditions: Dataset contains multiple features with different units or ranges.; The model is sensitive to feature magnitudes (e.g., distance-based or gradient-based).

Import `MinMaxScaler` from `sklearn.preprocessing`, instantiate it, and call `fit_transform` on the selected DataFrame columns. This fits the scaler to the data (learning min/max) and transforms the values to the [0, 1] range.

Conditions: Input data is numerical.; Scikit-learn library is installed.

Standardization transforms a feature value $x$ into $z = \frac{x - \mu}{\sigma}$, targeting a mean of 0 and a standard deviation of 1 for the transformed feature.

Conditions: Applied per numerical feature.; Uses the feature mean $\mu$ and standard deviation $\sigma$.; Assumes $\sigma > 0$.

Import `StandardScaler` from `sklearn.preprocessing`, instantiate it, and call `fit_transform` on the selected DataFrame columns. This transforms the data to have mean 0 and standard deviation 1.

Conditions: Input data is numerical.; Scikit-learn library is installed.

Min-max normalization transforms a feature value $X$ into $X_{\text{norm}} = \frac{X - X_{\min}}{X_{\max} - X_{\min}}$, mapping the feature's observed range onto the interval $[0, 1]$.

Conditions: Applied per numerical feature.; Requires a finite minimum $X_{\min}$ and maximum $X_{\max}$ for the feature.; Assumes $X_{\max} \neq X_{min}$ to avoid division by zero.